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Math Help - rational expression

  1. #1
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    rational expression

    getting some trouble
    with dis

    square root of x^2+6x=x+square root of 2x
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  2. #2
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    Quote Originally Posted by math321 View Post
    getting some trouble
    with dis

    square root of x^2+6x=x+square root of 2x
    square both sides ...

    x^2 + 6x = x^2 + 2x\sqrt{2x} + 2x

    combine like terms ...

    4x - 2x\sqrt{2x} = 0

    factor ...

    2x(2 - \sqrt{2x}) = 0

    set each factor equal to 0 and solve ... check solutions in the original equation.
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  3. #3
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    Quote Originally Posted by math321 View Post
    getting some trouble
    with dis

    square root of x^2+6x=x+square root of 2x
    \sqrt{x^2+6x}=x+\sqrt{2x}

    \sqrt{x^2+6x}=\sqrt{\left(x+\sqrt{2x}\right)^2}

    \sqrt{x^2+6x}=\sqrt{x^2+2x\sqrt{2x}+2x}

    6x=2x\sqrt{2x}+2x

    6x-2x=4x=2x\sqrt{2x}

    Do not do the following....

    \frac{4x}{2x}=2=\sqrt{2x}

    2=\sqrt{4}=\sqrt{2x}

    as you lose a solution.

    Instead, factor...

    4x-2x\sqrt{2x}=0

    x\left(4-2\sqrt{2x}\right)=0
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  4. #4
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    wondering what will be wrong in solving algebraically to find x which eventually equals 2? as problem is neither a quadratic equation nor an inequality?
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  5. #5
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    Quote Originally Posted by math321 View Post
    getting some trouble
    with dis

    square root of x^2+6x=x+square root of 2x
    Please use brackets to make you equation unambiguous, the above could plausibly mean either:

    square_root(x^2+6x)=x+square_root(2x)

    or

    square_root(x^2)+6x=x+square_root(2x)

    I know that the first of these is the most likely, but someone will always fail to make that assumption and interpret it differently.

    CB
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  6. #6
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    Quote Originally Posted by hotspice View Post
    wondering what will be wrong in solving algebraically to find x which eventually equals 2? as problem is neither a quadratic equation nor an inequality?
    It finds the x=2 solution, but not the x=0 solution.
    Quadratic equations may be solved by factoring, but factoring when possible, finds all solutions and is not confined to quadratics.

    2x=x\sqrt{2x}

    x\left(2-\sqrt{2x}\right)=0

    zero multiplied by another value is zero, so the factored version shows that x=0 is also a solution.

    If we divide both sides by x, we lose that particular solution, though it finds the x=2 answer.
    Also remember that we really should not divide by x because we cannot divide by zero anyway,
    so when we do that, we are finding only non-zero solutions.

    Do check CaptainBlack's post also.
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